View metadata, citation and similar papers at core.ac.uk brought to you by CORE provided by Elsevier - Publisher Connector A Dominant Negative Eigenvalue of a Matrix of Redheffer Tyler J. Jarvis* Department of Mathematics Bingharn Young University Provo, Utah 84602 Submitted by Richard A. Brualdi ABSTRACT Define the matrix A, as the sum of D,, =(dij) and C,, where dij = 1 when i 1j and zero when i + j, and where C, = (0,l,l,. , l)r(l,O,O,. .. ,O). We use the GerSgorin disc theorem and previous results regarding the spectral radius and the characteristic polynominal of A,, to prove the existence of a real eigenvalue which is asymptotically equal to - A as n + m. We also show that all eigenvalues, except the spectral radius and eigenvalue of order - 6, are (for large n> contained in the circle {z==(Iz--l<p(A,)/l o g n 1, w h ere p(A,) is the spectral radius. 1. INTRODUCTION In 1977, Redheffer [S] introduced a matrix which is closely connected with Merten’s function and hence with the Riemann hypothesis. The connec- tion is as follows: Redheffer’s matrix A, (A, is actually the transpose of Redheffer’s matrix) is defined as A,, = D,, + C,, where D,, = (dij), 1 if ilj, dij = 0 if i+j. *This research was supported in part by summer research grants in 1988 and 1989 from the Brigham Young University College of Physical and Mathematical Sciences. LZNEAR ALGEBRA AND ITSAPPLZCATZONS 142:141-152 (1990) 141 0 Elsevier Science Publishing Co., Inc., 1990 655 Avenue of the Americas, New York, NY 10010 0024-3795/90/$3.50 142 TYLER J. JARVIS and C, = (0, 1, 1, 1, . , l)T(l, O,O, . ,O). For example ‘1 1 1 1 1 1’ 110101 A,= 1 0 1 0 0 1 h 10 0 10 0’ 100010 \l 0 0 0 0 I/ Now if p :Zf-+ { - l,O, 1) is the Mobius function, and M : Zt -+ 2 is Merten’s function M(n) = t /J(k), k=l then (see [5]) detA,,=M(n). Hence, the Riemann hypothesis is true if and only if ldet A,( = 0( ni+‘) for every & > 0. Compare, for example, Titchmarsh [6] or IviE [3]. In [l], Barrett, Forcade, and Pollington found upper bounds for the coefficients of the characteristic polynominal of A,, and using these bounds, they proved that the Perron root of A,,, p(A,), is asymptotically equal to 6 as n goes to infinity. In this paper we are concerned with the remaining eigenvalues of A,. Barrett has made several interesting conjectures based on numerical evidence regarding these eigenvalues. A summary of these conjec- tures can be found in [4]. Prominent among these conjectures was that A,, has a negative (real) eigenvalue of magnitude - 6, and that the remaining eigenvalues are bounded close to the origin. In Theorem 2 we prove the existence of this negative real eigenvalue, that it is asymptotically equal to - 6 as n + m, and that the remaining eigenvalues have magnitude O(&/ log n). DOMINANT NEGATIVE EIGENVALUE 143 2. THE POLYNOMINAL g,(x) Barrett et al. [l] proved that the characteristic polynominal of A,, is given by det(tZ-A,,)=(t-l)“-“-‘f,,(t-l), where s = [log, n] (th e notation [xl means, as is usual, the largest integer less than or equal to x) and jn(x) = p+l _ e v,kXs-k k=l Here the coefficients are V nl =n-1, (1) and all of the coefficients onk are positive and bounded above by (logn)“_’ vflk<n (A-I)! (2) Hence if r is a root of f,,(x), then r + 1 is an eigenvalue of A,, and all eigenvalues A of A,, that do not satisfy f,,(h - 1) = 0 are identically equal to 1. We also need a result from [l] on the spectral radius of A,,, namely, if x, + 1 is the spectral radius of A,, then x, is a root of f,(x) and Consider now the polynominal g,(r)=f”o=x”+a,,r”-~+a,2x”-2+ -** +uns. x - *n 144 TYLER J. JARVIS Here, again, s = [log, n]. We note for later use that the first three coefficients are (by synthetic division) a ni = Xn, (3.1) =x 2 -v a n2 ” “1) (3.2) a “3 = ix; - X,V”, - vn2. (3.3) To obtain a useful general formula for an4 we work backward through the synthetic division to get a,,=)n,->O x,, and 1 ana-k = -(ans-k+l +0,*-k) >o for O<k<s. 2, The coefficients ank are asymptotically bounded above, as given in Theorem 1. THEOREM 1. If (Y, = maxr c k Q s{ a,& ) then 0 < (Y, = O<x,W log n), where 2 w=l+- < 3.8854. log2 The proof is by induction. First, by considering the definition of z),k, called c(n, k + 1) in [l], it follows’ that v,, < s + 1 =Glog, n + 1. Hence log, n + 1 log n an8 < =o - =O(x,“logn). x7l i 6 1 ‘One way to see this fact is by making lists of successive divisors 1, 1,, 1,, I,, , I, of length k + 1, such that Zi I Zi+1, Ii < Zi+,, and 2, < n. Each such list corresponds to one distinct cycle of length k + 1 in the graph G in [l]. Th e number of such lists is the coefficient tank. The list of length k + 1 with the smallest value of 1, is the list 1,2,4,. , 2k. The next lowest values of Zk is 3~2~-‘, and there are k distinct lists that allow 3X2’-’ as the last term. If k = [log, nl, then these k + 1 different lists are the only possibilities and I),~ < k + 1. DOMINANT NEGATIVE EIGENVALUE 145 Also, from equation (3) we have anj = O(x,W log n) for 1~ j < 3. Now for Q<m<k<s--4wehavebyhypothesis an- = O( x,” log n) and u,,~_.~ = so From Equation (2) it is clear that vns-k n(logn)“-k-’ -=G X” x,(s-k-l)! (logn)bx~+k-l ([log, n] - k -l)! Let z+l=log,n-k-l, so that 2t+2+k=n and 2<.z<log,n-3. Stirling’s formula gives n! > nnel-“\ln/2, whence n (logn)“+’ fllogn elogn ’ <--- i x, 1zei-$pBT i z I e(z+2+k)log2 =- z (elog2)’ < -ek+‘+(nlFfl n), ” 146 TYLER J. JARVIS where 4(n) = (e log2)” = (e log2)‘og2n-k-2. But 1+loglog2 = logn-(1+loglog2)(k +2). log2 SO n(l + log log 2)/ log 2 += e(l+loalog2Xk+2) ’ which gives nlog n k+2 nu+‘%‘@)/‘~~2 V ns-k -G-e ,(k+2)(1+140g2) x73 XII nlogn G-n (1 +loglog2)/ log2 e(k +2x -1oglog2) X” Let rCr(k)=e(k+2X-loglog2) Then log+(k)=(k+2)(-loglog2), andsinceO<k+2<s-2<log,n,weget - log log 2 log+(k) < (log n) log2 ) * < n - (log’% a/ 1% 2 DOMINANT NEGATIVE EIGENVALUE 147 From these results we have nlogn v,s-k _,l/h2 = r?l and hence ens-k = o(r;+2”0g2)log n. 3. ROOTS OF g,(x) We now resolve the conjecture mentioned above by showing that g,(r) has a negative real root of order - 6. Th eorem 2 guarantees that this is always the case and that all remaining roots have modulus less than or equal to X, /log n. THEOREM 2. For large n, g,(r) has a real root y, such that - xn < yn < - x, f O(log5 nX;-“), where w = 1 + 2/lag 2. Furthermore, all renlaining roots of g,(r) are con- tained in the disc The proof depends on GerSgorin’s disc theorem. First, consider the companion matrix Ce,(xj of g,(r): the s x s matrix 0 0 0 ... 0 -a,, -I 1 0 0 ... 0 -ans_l 0 1 0 ... 0 -ans-2 G r,(x) = 0 0 1 **. 0 -ans_3 ’ . 0 ... 0 0 1 -a,, 148 TYLER J. JARVIS Now conjugate c,~ with P = diag[p,, ~2,. -., ~~1,Pi > 0, to EFt . P. 0 0 0 0 - arts--- Pl Pl . Ps - 0 0 0 -ans-I--- P?. P2 P2 Ps 0 0 * * . 0 - a,,_,- P- ‘C&P = ii P3 PS _P3 *.. 0 0 0 - ans-3- P4 P4 Pi-1 Ps . ‘ . 0 0 0 -X,6--- Ps P,S The GerSgorin (TOW)disc centered at - x, is disjoint from the remaining discs if the radius of every disc centered at zero is less than x, - ps_ t /p,. Choose P such that the radii of all discs centered at zero are equal, i.e. Ps Pk P, _=- a + ans-k- Vk E(l,...,S_2}, ns Pi pk+l pktl and let t = p, /ansps. It foIIows immediately that Pk+l= t(Pk + ans-kPs) = tk( PI + ans-lPs) -t tk-la,,s-pPs + tko2ans_3Ps + * ’ - + b,s-kPst and that Ps-I t?J1 + ts_2a -= ns_, + tse3ans-2 + - * * + tan2 PS PS = tsmlans + tST2 arts-1 + * - * + ta,,. DOMINANT NEGATIVE EIGENVALUE 149 The discs are disjoint if 1 P,-1 _ ts-la _ . - ta,, . tcX”--=rPs n Its In other words, if we let T,(t) = tSans + ts-lans_l + . * + t2an2 - tx, + 1, then the discs are disjoint if T,(t) < 0 for some choice of t > 0. A choice of t E (0,11 gives T,(t) < t4 t ank + Pan3 + t2an2 - x,t + 1 k=4 < t4(log2 n -3) 4ykaasa,k + t3an3 + t2an2 - x,t + 1.